Solve $\left(\frac mn\right)^k=0.\overline{x_1x_2...x_9}$ (no computers!)

You want to find whether there exists a prime $p$ such that $p^2\mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that $$p\leq \sqrt{n}<578.$$ It is easily seen that $p>11$, so $$13\leq p\leq 577.$$ However, if $p>67$, then $$\frac{n}{p^2}\leq \frac{n}{71^2}<66.$$ Thus, $n$ must have a prime divisor $q<66$ such that $q\mid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3\cdot 37$. This leaves $13$ primes to deal with.

There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.

To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $\sqrt[3]{333667} \approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.